LINEAR OPERATORS OF BOUNDED ON SEMI-UNITARY SPACES
Let U be a finite-dimensional vector space over the field C. The vector space U is a semi-unitary space with semi-inner product [·, ·] if and only if U is an inner product space with inner product ?·, ·? and there exists a A linear operator on U which is positive semi-definite such that for every...
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Format: | Theses |
Language: | Indonesia |
Online Access: | https://digilib.itb.ac.id/gdl/view/73492 |
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Institution: | Institut Teknologi Bandung |
Language: | Indonesia |
Summary: | Let U be a finite-dimensional vector space over the field C. The vector space U is a
semi-unitary space with semi-inner product [·, ·] if and only if U is an inner product
space with inner product ?·, ·? and there exists a A linear operator on U which is
positive semi-definite such that for every x, y ? U, it holds [x, y] = ?Ax, y?. In
this thesis, a bounded linear operator on semi-unitary spaces will be studied where
the sufficient and necessary conditions of the linear operator A is bounded (i.e.
?Au? ? c?u? for a positive number c ? R and for all u ? U) is the isotropic
subspace U0 is A-invariant.
Suppose that T is a linear operator where U is a complex semi-inner product space,
the adjoint linear operator of T is defined as the linear operator S which satisfies
the property that for every x, y ? U, [Tx, y] = [x, Sy] holds. However, the adjoin
operator of T does not necessarily exist. Even if there is an adjoin operator, the
adjoin operator is not necessarily unique. Therefore, the sufficient and necessary
condition for the existence of an adjoin operator is the linear operator being bounded.
Suppose the set of all bounded linear operators B(U) that form a nonempty set, so
the set of bounded operators B(U) forms a vector space. Since the set B(U) forms a
vector space, This vector space has subspaces, one example of a subspace of B(U)
is S = {f : U ? U | f a bounded linear operators of U with f(U) ? U0}. |
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